**4.7 Inverse Trigonometric Functions**

**I. The inverse sine function**:

y = arc sin x if and only if sin y = x

We will use a capital A for Arcsin to show when the range is restricted so therefore:

y = Arc sin x

Domain: [-1, 1]

Range: restricted to [-π/2, π/2 ]

y = sin x some points are {(- π/2, -1), (0,0), (π/2, 1)}

so for the inverse, switch the x-values and y-values to get

y = Arc sin x some points are {(-1, - π/2), (0, 0), (1, π/2)}

Sketch the graph

**Evaluating the inverse sine function**:

Example 1: Arc sin (-1/2) = x

x = {-30º, 330º, 210º, - π/6, 11π/6, 7 π/6}

Example 2: Arc sin √(2)/2 = x

x = {45º, 135º, π/4, 3 π/4}

Example 3: Arc sin -√(3)/2 = x

x = {-60º, -120º, 300º, 240º, -π/3, -2π/3, 5π/3, 4π/3}

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**II. The inverse cosine function**:

y = arc cos x if and only if cos y = x

Again, to show a restricted range:

y = Arc cos x

Domain: [-1, 1]

Range: [0, π]

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**II. The inverse tangent function**:

y = Arc tan x if and only if tan y = x

Domain: all the real numbers

Range: (-π/2, π/2)

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Recall f ( f

^{-1}(x)) = x and f

^{-1}(f (x)) = x

Therefore: sin (Arc sin x) = x

Arc sin (sin x) = x

cos (Arc cos x) = x

Arc cos (cos x) = x

tan (Arc tan x) = x

Arc tan (tan x) = x

This only applies with sine in the interval [-π/2, π/2]

This only applies with cosine in the interval [0, π]

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Example 1: sin (Arc sin 0.7) = 0.7

**Example 2**: tan (Arc tan 35) = 35

**Example 3**: cos [ Arc cos (-0.3)] = -0.3

**Example 4**: Arc sin (sin 3 π)

Recall 3 π is not in the interval [0, π] so sin (3π) = sin (π) = 0 so

Arc sin (0) in the interval [-π/2, π/2] is equal to 0.

**Example 5**: sin (Arc tan 4/3) =

Using Pythagorean Theorem and since tangent is opposite divided by adjacent. Therefore 4 is the opposite and 3 is the adjacent to the angle. Using Pythagorean Theorem, that means that the hypotenuse is 5. So the sine would equal 4/5.

**Example 6**: tan (arc cos x/5) = θ

This means that x is the adjacent and 5 is the hypotenuse.

a

^{2}+ b

^{2}= c

^{2}

x

^{2}+ b

^{2}= 5

^{2}

b

^{2}= 25 - x

^{2}

b = the square root of (25 - x

^{2})

so the tan θ = (the square root of (25 - x

^{2}))/5

**Example 7**: cot (arc tan 1/x)

This means that the tangent is equal to 1/x ,

tangent/1 = 1/x

taking the reciprocal of both sides gives us

1/tangent = x/1

1/tangent is equal to cotangent so

so the cotangent is equal to x.

**Example 8**: Arc cos 3/ (the square root of (x

^{2}- 2x + 10)) = Arc sin ___

Using Pythagorean Theorem:

3

^{2}+ b

^{2}= (the square root of (x

^{2}- 2x + 10))

^{2}

9 + b

^{2}= x

^{2}- 2x + 10

b

^{2}= x

^{2}- 2x + 1

b

^{2}= (x - 1)(x - 1)

b

^{2}= (x - 1)

^{2}

b = x - 1

Therefore sin θ = (x - 1)/(the square root of (x

^{2}- 2x + 10))

so the Arc cos 3/(the square root of (x

^{2}- 2x + 10)) = Arc sin ((x - 1)/(the square root of (x

^{2}- 2x + 10))

**Prove the identity**:

Arc sin x + Arc cos x = π/2

Let θ = Arc sin x and β = Arc cos x

By substitution: θ + β = π/2

Now since: Arc sin x = θ and β = Arc cos x

This means sin θ = x and cos β = x.

Since x = x, then sin θ = cos β

For this to happen, the triangle has to be a right triangle.

Therefore θ + β = π/2

Since this is the same answer as our substitution above, we now know

Arc sin x + Arc cos x = π/2

4.7 homework #34; pg. 351; #1 - 15 odd, 19 - 23 odd, 27 - 53 odd, 57, 59, 71, 87, 88

pg. 341; #19, 45, 52, 64, 91

Quiz on Sections 4.5 - 4.6 next class